A charge `-2muC` at the origin, `-1muC` at `+7cm` and `1muC` at -7cm are placed on X-axis. The mutual potential energy of the system is

Options

A. `-0.051J`
B. `-0.045J`
C. `0.045J`
D. `-0.064J`

Explanation

To find the mutual potential energy of the system of charges, we will follow these steps:

### Step 1: Identify the charges and their positions
- Charge \( Q_1 = -2 \, \mu C \) at position \( x_1 = 0 \, cm \) (origin).
- Charge \( Q_2 = -1 \, \mu C \) at position \( x_2 = +7 \, cm \).
- Charge \( Q_3 = +1 \, \mu C \) at position \( x_3 = -7 \, cm \).

### Step 2: Calculate the potential energy between each pair of charges
The formula for the potential energy \( U \) between two point charges is given by:
\[ U = k \frac{Q_1 Q_2}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, Nm^2/C^2 \) and \( r \) is the distance between the charges.

#### Pair 1: \( Q_1 \) and \( Q_2 \)
- Distance \( r_{12} = |0 - 7| = 7 \, cm = 0.07 \, m \)
- Potential energy \( U_{12} = k \frac{(-2 \times 10^{-6})(-1 \times 10^{-6})}{0.07} \)

Calculating \( U_{12} \):
\[ U_{12} = 9 \times 10^9 \frac{(2 \times 10^{-6})(1 \times 10^{-6})}{0.07} = 9 \times 10^9 \frac{2 \times 10^{-12}}{0.07} \]
\[ U_{12} = 9 \times 10^9 \times \frac{2}{0.07} \times 10^{-12} \approx 2.57 \times 10^{-2} \, J \]

#### Pair 2: \( Q_1 \) and \( Q_3 \)
- Distance \( r_{13} = |0 - (-7)| = 7 \, cm = 0.07 \, m \)
- Potential energy \( U_{13} = k \frac{(-2 \times 10^{-6})(1 \times 10^{-6})}{0.07} \)

Calculating \( U_{13} \):
\[ U_{13} = 9 \times 10^9 \frac{(-2 \times 10^{-6})(1 \times 10^{-6})}{0.07} = -2.57 \times 10^{-2} \, J \]

#### Pair 3: \( Q_2 \) and \( Q_3 \)
- Distance \( r_{23} = |7 - (-7)| = 14 \, cm = 0.14 \, m \)
- Potential energy \( U_{23} = k \frac{(-1 \times 10^{-6})(1 \times 10^{-6})}{0.14} \)

Calculating \( U_{23} \):
\[ U_{23} = 9 \times 10^9 \frac{(-1 \times 10^{-6})(1 \times 10^{-6})}{0.14} = -6.43 \times 10^{-3} \, J \]

### Step 3: Calculate the total potential energy of the system
Now, we sum up the potential energies of all pairs:
\[ U_{total} = U_{12} + U_{13} + U_{23} \]
\[ U_{total} = 2.57 \times 10^{-2} + (-2.57 \times 10^{-2}) + (-6.43 \times 10^{-3}) \]
\[ U_{total} = 0 - 6.43 \times 10^{-3} = -6.43 \times 10^{-3} \, J \]

### Final Answer:
The mutual potential energy of the system is:
\[ U_{total} = -6.43 \, mJ \]

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